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Inspired by this question Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$, I wonder if the following is true:

Let $k > 0$, and define $M_k(x_1,\ldots,x_n) = \left(\frac{1}{n}\sum_{i=1}^n x_i^k\right)^{1/k}$, and $M_0 = \left(x_1\cdots x_n\right)^{1/n}$. In the question it was proved that $$\tag{Q}\label{eq:Q}M_0(\mathbf{x})^{\frac{n}{n+2}}M_2(\mathbf{x})^{\frac{2}{n+2}}\leq M_1(\mathbf{x}),\qquad \mathbf{x}\succeq 0.$$ For brevity I'll omit $\mathbf{x}$ from the following.

Can we generalize $\eqref{eq:Q}$ to e.g. $$\tag{1}\label{eq:1} M_0^{\frac{n}{n+k}}M_k^{\frac{k}{n+k}} \leq M_1, $$ or perhaps even $$\tag{2}\label{eq:2} M_0^\theta M_t^{1-\theta}\leq M_1, $$ where $\theta \in (0,1), t = n(\theta^{-1}-1)$?

In a way this seems related to moment interpolation: if $\mu_t = \int_0^\infty x^tf(x)dx$ for some integrable function $f:[0,\infty]\mapsto[0,\infty]$, then we have the inequality $$ \mu_t\leq\mu_{t_0}^{1-\theta}\mu_{t_1}^\theta~,\qquad t = (1-\theta)t_0+\theta t_1, \theta \in[0,1], $$ which is proven using Hölder's inequality.

Whether $\eqref{eq:1}$ and $\eqref{eq:2}$ are true or not, I wonder if there are even "nicer" generalizations to $\eqref{eq:Q}$.

V.S.e.H.
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    Not quite what you ask for, but the https://en.wikipedia.org/wiki/Lehmer_mean inequlaity may be of interest. – Blitzer Aug 15 '22 at 14:13
  • did you look at the solution I wrote to the linked question? That method will certainly give you some version of your inequality but I am not sure if it will be $\dfrac{k}{n+k}$ weight on $M_k.$ It should hold when $k<<n$ in some appropriate sense though. – dezdichado Oct 08 '22 at 18:07
  • @dezdichado Yes, as I commented under your answer, it is a very nice generalization :) I suspect that we can generalize even further, by looking at this bound as a sort of "interpolation" between means? Alas, I sort of abandoned this question for awhile, will look at it more seriously soon. – V.S.e.H. Oct 08 '22 at 20:06
  • you are looking at the wrong answer then. If you use my method, then $\dfrac{k}{n+k}$ would be too large for when $k$ and $n$ are of the same order. But you can definitely derive an interpolation like you want using the same idea. – dezdichado Oct 08 '22 at 21:03
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    @dezdichado Ah I see, you're right. Your method is indeed very elegant (+1'd long time ago), I will give it a try and play around with generalizations. – V.S.e.H. Oct 09 '22 at 14:16

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Let $$M_s(x)=\left(\frac{\sum_{i=1}^n x_i^s}{n}\right)^{1/s}$$ and $M_0(x)$ be the corresponding geometric mean. It is well known that the mean $M_s(x)$ is increasing in $s$ for $-\infty<s<\infty,$ and for positive $n-$tuples $(x_1,\ldots,x_n).$

In Mitrinovic's Analytic Inequalities, published by Springer many years ago in the Grundlehren series, one finds, on page 79, the following inequality on the ratio $Q_{s,t}(x)=M_s(x)/M_t(x)$ of means of order $-\infty<t<s<\infty:$ $$ Q_{s,t}(x)\leq \left( \frac{t(C^s-C^t)}{ (s-t)(C^t-1) } \right)^{1/s} \left( \frac{s(C^t-C^s)}{ (t-s)(C^s-1) } \right)^{-1/t},\quad st\neq 0, $$ and $$ Q_{s,t}(x)\leq \left( \frac{C^{t/(C^t-1)}}{ e \log(C^{t/(C^t-1)}) } \right)^{-1/t}, \quad s= 0, $$ where $$C=\frac{\max_i x_i }{\min_i x_i}.$$ I believe careful use of these inequalities can yield some conditions under which your desired results would hold, or maybe rule them out.

kodlu
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  • Interesting, thanks for the reference (incidentally I happen to have this book). But I don't see how this helps us. I'll try to prove the inequality using induction, as is done in the original question. I think however that the inequality is not true, since if we take $k\to\infty$ then we approach $M_\infty \leq M_1$, which is false. – V.S.e.H. Aug 26 '22 at 20:27